A non -uniform rod of length l having mass density `lamda(x) = (A+Bx^2)` is placed - x - axis with its ends at , `(a,0) and (a+ l, 0)` . The force it would exert on a point mass m kept at the origin is

Options

A. `Gm[A(l/(a(a+2)))+Bl]`
B. `Gm[A(l/(a(a-l)))-Bl]`
C. `Gm[A(l/(a(a+l)))+Bl]`
D. `Gm[A(l/(a(a+l)))-Bl]`

Answer

To find the force exerted by a non-uniform rod on a point mass \( m \) placed at the origin, we will follow these steps:

### Step 1: Define the mass density of the rod
The mass density of the rod is given by:
\[ \lambda(x) = A + Bx^2 \]
where \( A \) and \( B \) are constants.

### Step 2: Determine the mass of an infinitesimal element of the rod
Consider an infinitesimal element of the rod at position \( x \) with a length \( dx \). The mass \( dm \) of this element can be expressed as:
\[ dm = \lambda(x) \, dx = (A + Bx^2) \, dx \]

### Step 3: Calculate the gravitational force exerted by the infinitesimal element on the point mass
The gravitational force \( dF \) exerted by the mass \( dm \) on the point mass \( m \) located at the origin is given by Newton's law of gravitation:
\[ dF = \frac{G \, dm \, m}{r^2} \]
where \( r \) is the distance from the element \( dm \) to the point mass \( m \). Since the point mass is at the origin and the element is at position \( x \), we have \( r = x \). Thus,
\[ dF = \frac{G \, (A + Bx^2) \, dx \, m}{x^2} \]

### Step 4: Express the force in terms of \( dx \)
We can simplify \( dF \):
\[ dF = Gm \left( \frac{A + Bx^2}{x^2} \right) dx = Gm \left( \frac{A}{x^2} + B \right) dx \]

### Step 5: Integrate to find the total force exerted by the rod
To find the total force \( F \) exerted by the entire rod on the point mass, we need to integrate \( dF \) from \( x = a \) to \( x = a + l \):
\[ F = \int_{a}^{a+l} dF = \int_{a}^{a+l} Gm \left( \frac{A}{x^2} + B \right) dx \]
This can be separated into two integrals:
\[ F = Gm \left( \int_{a}^{a+l} \frac{A}{x^2} \, dx + \int_{a}^{a+l} B \, dx \right) \]

### Step 6: Calculate the integrals
1. The first integral:
\[ \int \frac{A}{x^2} \, dx = -\frac{A}{x} \]
Evaluating from \( a \) to \( a+l \):
\[ \left[-\frac{A}{x}\right]_{a}^{a+l} = -\frac{A}{a+l} + \frac{A}{a} = A \left( \frac{1}{a} - \frac{1}{a+l} \right) = A \left( \frac{(a+l) - a}{a(a+l)} \right) = \frac{Al}{a(a+l)} \]

2. The second integral:
\[ \int B \, dx = Bx \]
Evaluating from \( a \) to \( a+l \):
\[ [Bx]_{a}^{a+l} = B(a+l) - Ba = Bl \]

### Step 7: Combine the results
Now substituting back into the expression for \( F \):
\[ F = Gm \left( \frac{Al}{a(a+l)} + Bl \right) \]
Factoring out \( l \):
\[ F = Gml \left( \frac{A}{a(a+l)} + B \right) \]

### Final Result
Thus, the force exerted by the rod on the point mass \( m \) at the origin is:
\[ F = Gml \left( \frac{A}{a(a+l)} + B \right) \]